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Functions whose differentials are linearly independent at the point in question. For $F=(F_1,\dots ,F_r)$ this means the derivative has rank $r$, so $F$ is a submersion there. The condition is open. It is distinct from global algebraic independence of functions.

Let $P_w(E_4,E_6)$ be weighted homogeneous of modular weight $w$, and suppose ${\sum }_w{P}_w(z)=0$ as a holomorphic function. Apply the modular group matrices $\left(\begin{array}{cc}1& 0\\ n& 1\end{array}\right)$. At a fixed point $z$ in the complex upper half-plane, this gives ${\sum }_w(nz+1{)}^w{P}_w(z)=0$ for every integer $n$. A polynomial with infinitely many roots vanishes identically, so every $P_w(z)=0$. Within one modular weight, the monomials reduce to powers of $E_4^3/E_6^2$, after taking out a common factor. This ratio is nonconstant, as its Fourier expansion begins $1+1728q+\cdots$, so its powers are linearly independent. This proves algebraic independence even when the generators are regarded as analytic functions rather than elements of a formal graded ring.

A standard form of the Schneider-Lang theorem is this. Let $f_1,\dots ,f_s$ be meromorphic functions with $f_j^{\prime }\in K[f_1,\dots ,f_s]$ for a fixed number field $K$. Suppose $f_1,f_2$ have algebraic independence over $\mathbb{C}$ and have finite growth orders ${\rho }_1,{\rho }_2$. There are at most $[K:\mathbb{Q}]({\rho }_1+{\rho }_2)$ points at which all the functions are finite and all their values lie in $K$.

Take $f_1(z)=z$ and $f_2(z)=e^z$, of orders zero and one. Their derivatives belong to $K[z,e^z]$. They have algebraic independence: a polynomial relation ${\sum }_j{p}_j(z)e^{jz}=0$ is impossible by exponential-polynomial independence, or by letting real $z$ tend to infinity and considering the largest $j$.

If nonzero algebraic $\alpha$ had algebraic $e^{\alpha }$, put $K=\mathbb{Q}(\alpha ,e^{\alpha })$. At each of the distinct points $z=n\alpha$, both $z$ and $e^z=(e^{\alpha }{)}^n$ belong to $K$. Infinitely many points contradict the theorem's finite bound. Therefore

In Siegel's original convention, an E-function is an entire series $F(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$ with coefficients in a fixed number field, satisfying a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$, together with these arithmetic bounds: for every $ϵ>0$, all conjugates of $a_0,\dots ,a_n$ have size $O_{ϵ}(n^{ϵn})$, and some positive integer $d_n=O_{ϵ}(n^{ϵn})$ makes every $d_n{a}_j$, $j\le n$, an algebraic integer. A frequently used stricter convention replaces both bounds by $C^{n+1}$. The original and strict growth definitions should not simply be declared equivalent; the exponential examples below satisfy the strict one.

For sums, work in the compositum of the coefficient fields. Coefficient conjugate bounds add. The product of the two common denominators clears both lists; using $ϵ/2$ in each original denominator bound gives the required $O_{ϵ}(n^{ϵn})$ product bound. Finally all derivatives of $F+G$ lie in the sum of the two finite-dimensional derivative spaces over that rational function field, so they satisfy a nontrivial linear dependence, giving a differential equation for the sum. Thus sums are again E-functions, the addition closure of Siegel E-functions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $F^{\prime }=A(z)F$ with rational-function matrix over $\overline{\mathbb{Q}}$, then at every nonzero algebraic $\xi$ which is not a pole of $A$,In particular algebraically independent functions give algebraically independent values.

Take $F_j(z)=e^{{\gamma }_{j}z}$ for algebraic numbers ${\gamma }_j$ linearly independent over $\mathbb{Q}$. Their coefficients are ${\gamma }_j^n$ with exponential conjugate and denominator bounds, and $F_j^{\prime }={\gamma }_j{F}_j$. A polynomial relation among the functions expands into a relation among exponentials with distinct exponents $\sum m_j{\gamma }_j$ and polynomial coefficients after clearing rational-function denominators; exponential-polynomial independence excludes it. At $\xi =1$ the theorem gives algebraic independence of the $e^{{\gamma }_j}$. For arbitrary distinct algebraic ${\alpha }_i$, choose a rational basis of their span and divide that basis by a common denominator. The $e^{{\alpha }_i}$ are distinct Laurent monomials in the corresponding algebraically independent exponential values, so are linearly independent over $\overline{\mathbb{Q}}$. This recovers the general Lindemann–Weierstrass theorem. Its simplest instance uses just $F(z)=e^z$ at nonzero algebraic $\xi$.

Normalize the weight-four and weight-six Eisenstein series to have constant term one. Their Fourier expansions beginTheir absolutely convergent lattice sums transform by modular weights four and six under reindexing by the modular group, and their Fourier expansions prove holomorphy at its modular cusp. Thus they are elements of $M_4,M_6$.

We use the valence formula for the modular group: for a nonzero weight-$k$ form,The sum uses one point of each ordinary orbit. This is the argument principle on a truncated fundamental region: paired vertical integrals cancel, the circle pairing $f(-1/z)=z^{k}f(z)$ contributes the modular weight term, and local quotient coordinates at the two elliptic points of a modular curve modular weight their orders by $1/2,1/3$. All orders are nonnegative for a holomorphic modular form.

Define the modular discriminant byIt is a weight-twelve cusp form. Its order at infinity is one, exhausting the valence total $12/12=1$. Hence $\Delta$ has no zeros in $\mathbb{H}$. Therefore every weight-$k$ cusp form is divisible by $\Delta$, with quotient holomorphic on $\mathbb{H}$ and at the modular cusp:Negative modular weights vanish by the valence formula, and odd modular weights vanish because $-I$ acts by $(-1{)}^k$. Weight-zero forms are constant: they descend holomorphically to the compactified modular curve, and the maximum modulus principle applies. There are no weight-two forms: $f(i)=i^{2}f(i)=-f(i)$ forces a zero at $i$, whose valence contribution $1/2$ exceeds $2/12$.

Every nonnegative even modular weight other than two is $4a+6b$ with $a,b\ge 0$. If $k\equiv 0(\mathrm{mod}4)$, take $b=0$; if $k\equiv 2(\mathrm{mod}4)$ and $k\ge 6$, take $b=1$. For $f\in M_k$, choose such a monomial $h=E_4^a{E}_6^b$, whose constant term is one. Subtracting the constant term $c$ of $f$ gives $f-ch\in S_k$, soInduction on modular weight, together with $1728\Delta =E_4^3-E_6^2$, proves that every modular form is a polynomial in $E_4,E_6$.

For algebraic independence, first fix one modular weight $w$. All solutions of $4a+6b=w$ have the same residue of $a$ modulo three and the same parity of $b$. Consequently their monomials can be writtenfor suitable $a_0,b_0,L$. On an open set where the common factor and $E_6$ are nonzero, a linear relation would give a polynomial relation inThis is nonconstant. By the open mapping theorem, a polynomial vanishing on all its values must be the zero polynomial. Thus same-weight monomials are linearly independent.

Different modular weights cannot cancel either. Suppose ${\sum }_w{P}_w(E_4,E_6)=0$ with each $P_w$ weighted homogeneous. Transform by ${\gamma }_n=\left(\begin{array}{cc}1& 0\\ n& 1\end{array}\right)$. At any fixed $z\in \mathbb{H}$ this givesThe numbers $nz+1$ are infinitely many distinct values, so the polynomial in that number vanishes identically, and every $P_w(z)=0$. The same-weight independence then sets every coefficient to zero. This is the graded weights separate analytic polynomial relations argument. Hencewith algebraically independent generators. Multiplying the two Eisenstein series by their nonzero lattice-normalization constants does not alter either generation or independence.

Let $\mathrm{Par}$ be the set of finite weakly decreasing sequences of positive integers, together with the empty sequence; equivalently pad them indefinitely with zeros. The size is $|\lambda |={\sum }_i{\lambda }_i$, the length counts positive parts, and $\mathrm{Par}(n)$ consists of size-$n$ integer partitions. Their Young diagrams have cells $(i,j)$ with $1\le j\le {\lambda }_i$.

Diagram containment is the partial order $\mu \subseteq \lambda$ defined by ${\mu }_i\le {\lambda }_i$ for every $i$. Componentwise minimum and maximum give its meet and join, so this is Young's lattice. Its covers add one cell. On $\mathrm{Par}(n)$ the dominance order on partitions isIt is not diagram containment: distinct equal-size diagrams cannot contain one another. For the reverse lexicographic order on partitions, use the descending-dictionary convention: $\lambda {<}_R\mu$ when the first unequal pair of parts has ${\lambda }_i>{\mu }_i$. Thus for size four its increasing list is $(4),(3,1),(2,2),(2,1,1),(1,1,1,1)$. This is a total order opposite to increasing dictionary order on integer partitions, not the last-differing-exponent order on polynomial monomials.

Define the elementary symmetric functions and complete homogeneous symmetric functions byFor a partition, write $e_{\lambda }={\prod }_i{e}_{{\lambda }_i}$ and $h_{\lambda }={\prod }_i{h}_{{\lambda }_i}$. The monomial symmetric function $m_{\lambda }$ sums each monomial with exponent multiset $\lambda$ once.

Here is a proof of the Fundamental theorem of symmetric polynomials sufficient for the stable algebra. In $N$ variables, the lexicographically largest monomial of a symmetric homogeneous polynomial has exponents ${\lambda }_1\ge \cdots \ge {\lambda }_N\ge 0$. Otherwise exchanging an inverted pair of variables produces a larger monomial with the same coefficient. The polynomialhas that monomial as its largest one, with coefficient one. Subtract a suitable multiple and continue; there are only finitely many monomials of the fixed degree, so the procedure terminates and proves generation. Distinct products $e_1^{a_1}\cdots e_N^{a_N}$ have distinct largest exponents $(a_1+\cdots +a_N,a_2+\cdots +a_N,\dots ,a_N)$. The largest such monomial in any nonzero polynomial relation cannot cancel, proving algebraic independence. Stabilizing in each degree, where only $e_r$ with $r$ no greater than that degree can occur, gives

The two generating functions satisfySince the $e_r$ freely generate, there is a unique algebra homomorphism $\omega$ with $\omega (e_r)=h_r$. Apply it to the last identity:It follows that $\omega (h_r)=e_r$ and ${\omega }^2(e_r)=e_r$. Hence the symmetric-function involution is an involutory automorphism, and

For the forgotten coefficients, it is useful to derive the required duality explicitly. Put $p_r={\sum }_i{x}_i^r$. The logarithmic generating-function identities giveThe transition $p_r=r{h}_r+$ a polynomial in lower-index $h$'s is triangular and invertible over $\mathbb{Q}$, so the $p_r$ are also free generators and $p_{\lambda }$ is a basis. Define the Hall inner product of symmetric functions byExpand the same formal kernel in two ways:The last expansion is a reproducing kernel for this inner product. Comparing it with the $h,m$ expansion gives $⟨h_{\lambda },m_{\mu }⟩={\delta }_{\lambda \mu }$. One can see this without assuming any Schur orthogonality: pair either kernel with $h_{\nu }(y)$; the reproducing property gives $h_{\nu }(x)$, and uniqueness of its $h$ expansion proves the duality. The signs in $\omega (p_{\lambda })$ show that $\omega$ is an isometry; together with ${\omega }^2=1$ this makes it self-adjoint.

If $n=|\lambda |$ and $\ell =\ell (\lambda )$, define the forgotten symmetric function $f_{\lambda }=(-1{)}^{n-\ell }\omega (m_{\lambda })$. Its monomial coefficient is consequentlyIt remains to compute this coefficient rather than merely invoke duality. Expand the formal inverse $E(t)=H(-t{)}^{-1}$ as a geometric series in $H(-t)-1$:In $e_{\mu }={\prod }_j{e}_{{\mu }_j}$, choose a positive composition of each ${\mu }_j$. A term is $h_{\lambda }$ precisely when the concatenated list is a distinct rearrangement $\alpha$ of the parts of $\lambda$. Its boundaries at the end of each chosen composition are exactly the prescribed prefix sums of $\mu$. Conversely such a rearrangement uniquely determines the compositions by cutting at those prefix sums. Every one of its signs is $(-1{)}^{\sum {\mu }_j-\ell }=(-1{)}^{n-\ell }$, canceled by the prefactor in $a_{\lambda \mu }$. ThusIdentical parts are not artificially distinguished. For example $f_{(2,1)}=m_{(2,1)}+2m_{(3)}$: only the order $(2,1)$ refines $(2,1)$, while both orders refine the single block $(3)$.

An integer version of Siegel lemma is as follows. For $M<N$ homogeneous linear equations $Az=0$, with integer matrix entries bounded in absolute value by $A_0\ge 1$, there is a nonzero integer solution satisfyingPut the displayed integer bound equal to $Q$ and evaluate the equations at every vector in $\{0,\dots ,Q{\}}^N$. There are $(Q+1{)}^N$ inputs and at most $(2N{A}_{0}Q+1{)}^M$ outputs. But $2N{A}_{0}Q+1\le 2N{A}_0(Q+1)$ and $(Q+1{)}^{N-M}>(2N{A}_0{)}^M$, so the number of inputs is larger. Two distinct inputs have identical outputs; their difference proves the claim. This is the pigeonhole proof of integer Siegel lemma. Rational coefficients are handled by clearing denominators. Over a number field of degree $d$, expanding in a rational basis gives at most $dM$ rational equations, so the same construction applies when $N>dM$, with the height of the resulting integer matrix controlling the bound.

Siegel's original arithmetic definition starts with an entire series $E(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$, with all $a_n$ in a fixed number field. For every $ϵ>0$ there must be positive integers $d_n$ such that $d_n{a}_0,\dots ,d_n{a}_n$ are algebraic integers, and both $d_n$ and the absolute values of every conjugate of $a_n$ are $O_{ϵ}(n^{ϵn})$. The growth condition ensures convergence everywhere. The stronger, now customary strict E-function definition uses bounds $C^{n+1}$ for a fixed $C$ and also includes the requirement of a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$. In the original formulation, that differential condition is explicitly part of the system hypothesis in the theorem below. These definitions should be distinguished; the exponentials used here satisfy even the stronger conditions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $f^{\prime }=A(z)f$, with $A(z)$ a matrix of rational functions over $\overline{\mathbb{Q}}$, then for every nonzero algebraic $\xi$ which is not a pole of $A$,In particular, algebraic independence of the functions implies algebraic independence of the values. The exclusion of zero is necessary: their Taylor values there are algebraic.

Take rationally independent algebraic ${\gamma }_1,\dots ,{\gamma }_s$ and $f_j(z)=e^{{\gamma }_{j}z}$. Their coefficients are ${\gamma }_j^n$, their conjugates grow exponentially, and powers of a fixed common denominator clear all coefficients through degree $n$. They satisfy the nonsingular diagonal system $f_j^{\prime }={\gamma }_j{f}_j$.

To check functional independence, a polynomial relation among these exponentials, after clearing rational-function denominators, would be ${\sum }_m{P}_m(z)e^{{\lambda }_{m}z}=0$ with distinct ${\lambda }_m={\sum }_j{m}_j{\gamma }_j$. Such an exponential polynomial cannot vanish identically. To isolate a nonzero term with exponent ${\lambda }_0$, apply ${\prod }_{\lambda \ne {\lambda }_0}(D-\lambda {)}^{\mathrm{deg}{P}_{\lambda }+1}$. All other terms vanish, while the remaining polynomial is acted on by operators $D+{\lambda }_0-\lambda$, each injective on polynomials because its nonzero constant preserves a nonzero leading coefficient. Thus the remaining term is nonzero, a contradiction.

The theorem at $\xi =1$ now makes $e^{{\gamma }_j}$ algebraically independent. For arbitrary distinct algebraic exponents ${\alpha }_i$, choose a rational basis of their span and divide its elements by a common denominator so that every ${\alpha }_i$ is an integer combination of the resulting ${\gamma }_j$. Then $e^{{\alpha }_i}$ are distinct Laurent monomials in $e^{{\gamma }_j}$. A linear relation among them becomes a nonzero polynomial relation after multiplying by a common monomial, contradicting independence. These particular E-functions therefore yield the full Lindemann–Weierstrass theorem.

The coordinate ring is $\mathbb{C}[W]=\mathrm{Sym}(W^{\ast })$. The induced action is contragredient on functions:It is a group action by degree-preserving algebra automorphisms, extending the dual action on linear polynomial functions. Its polynomial invariant ring isFor a monic polynomial $f(t)={\prod }_{i=1}^d(t-r_i)$, its polynomial discriminant is $\mathrm{disc}(f)={\prod }_{i<j}(r_i-r_j{)}^2$. This is symmetric in the roots, hence polynomial in the coefficients, and is zero exactly when a root is repeated.

For the alternating group action, let $e_1,\dots ,e_n$ be the elementary symmetric polynomials and put $\Delta ={\prod }_{i<j}(X_i-X_j)$. If $p$ is $A_n$-invariant and $\tau$ is any transposition, decomposeNormality and index two of $A_n$ show that $p_{+}$ is symmetric and $p_{-}$ transforms by the sign character of $S_n$. Any alternating polynomial vanishes when $X_i=X_j$, so every $X_i-X_j$ divides it. These pairwise nonassociate prime factors have product $\Delta$, so $p_{-}=\Delta q$ with $q$ symmetric. The Fundamental theorem of symmetric polynomials yieldsThe sum is direct, since a polynomial that is both symmetric and alternating is zero in characteristic zero. Also ${\Delta }^2=D(e_1,\dots ,e_n)$, where $D$ is the discriminant polynomial ofThus, more precisely than the requested quotient assertion,Surjectivity follows from the direct-sum expression. Divide any putative kernel element by the monic quadratic in $Z$; its remainder is $a(T)+Zb(T)$. The direct sum forces $a(e)=b(e)=0$, and algebraic independence of the $e_i$ gives $a=b=0$. Hence the displayed relation is the entire kernel. The argument also covers $n=2$, where $A_2$ is trivial.

Now let $G$ be finite with no nontrivial linear characters, and write $R=\mathbb{C}[W]$, $S=R^G$. The polynomial ring $R$ is a unique factorization domain. Factor a nonzero invariant $f$ in $R$; invariance permutes the associate classes of its irreducible factors and makes their exponents constant on each orbit. For an orbit $\mathcal{O}$, choose representatives and form its orbit product of polynomial factorsFor every $g$, $g\cdot q_{\mathcal{O}}=\theta (g)q_{\mathcal{O}}$ for a nonzero scalar $\theta (g)$. Applying two group elements proves that $\theta :G\to {\mathbb{C}}^{\times }$ is a homomorphism. The hypothesis forces $\theta =1$, so $q_{\mathcal{O}}\in S$.

This orbit product is prime in $S$. If it divides $ab$ with $a,b\in S$, one factor $p\in \mathcal{O}$ divides $a$ or $b$ in $R$. Invariance of that chosen polynomial makes every factor in the orbit divide it. Their product therefore divides it in $R$, and the quotient is invariant because numerator and denominator are invariant and cancellation is valid in the integral domain $R$. Thus $q_{\mathcal{O}}$ divides $a$ or $b$ in $S$. Every nonzero $f\in S$ is a scalar times a product of these prime orbit products, and the only units of $S$ are nonzero constants, as they are units in $R$. Therefore $\mathbb{C}[W{]}^G$ is a unique factorization domain. This gives a direct proof, without assuming a divisor-class-group theorem.

For a failure in characteristic zero, let the cyclic group of order two act on ${\mathbb{C}}^2$ by $-I$. Its coordinate invariants areEvery invariant monomial has even total degree: its exponents are either both even, or both odd, giving the indicated generators. Reducing powers of $b$ to at most one shows that $ac-b^2$ is the only relation, since $a^i{c}^j$ and $b{a}^i{c}^j$ map to distinct monomials. The three quadratic invariants are irreducible in $S$: each nonconstant invariant has degree at least two, so a product of two nonunits has degree at least four. They are pairwise nonassociate, butgives two different irreducible factorizations. Hence this invariant ring is not a unique factorization domain. The nontrivial sign character is precisely the kind of character excluded in the preceding theorem.

The Noether normalization lemma is valid over every field $k$, including finite fields: a nonzero finitely generated algebra $C$ contains elements $y_1,\dots ,y_d$ with algebraic independence over $k$ such that $C$ is finite as a module over the polynomial ring $k[y_1,\dots ,y_d]$. We prove it without an infinite-field hypothesis.

Write $C=k[a_1,\dots ,a_n]$ and induct on $n$. If the generators have algebraic independence, they already give a polynomial ring and the assertion is immediate. Otherwise choose a nonzero relation $F(a_1,\dots ,a_n)=0$. Choose an integer $M$ larger than every exponent in the monomials of $F$, and setIn $F(Y_1+T^M,Y_2+T^{M^2},\dots ,Y_{n-1}+T^{M^{n-1}},T)$, the largest power of $T$ has a nonzero coefficient in $k$. Indeed, a monomial with exponents $(e_1,\dots ,e_n)$ contributes top weight $e_n+{\sum }_{i<n}{e}_i{M}^i$. These weights are distinct by uniqueness of base-$M$ expansion, so only one monomial contributes the highest power. After rescaling, the substituted relation is monic in $T$.

Consequently $a_n$ is an integral element over $C^{\prime }=k[b_1,\dots ,b_{n-1}]$, and $C=C^{\prime }[a_n]$ is finite over $C^{\prime }$. Apply the induction hypothesis to $C^{\prime }$ and compose the finite module extensions. This proves Noether normalization by weighted substitutions. The induction reaches $n=0$, where $C=k$. For an integral domain $C$, taking fraction fields makes the resulting extension finite algebraic, so $d$ is the transcendence degree of $\mathrm{Frac}(C)/k$. More generally $d=\dim C$: integral extensions preserve Krull dimension, and a polynomial algebra in $d$ variables has Krull dimension $d$.

A useful bridge to the Hilbert Nullstellensatz is the Zariski lemma. If a field $E$ is a finitely generated algebra over $k$, normalization makes it finite and integral over a polynomial ring $P=k[y_1,\dots ,y_d]$. A subring over which a field is integral is a field: for nonzero $a\in P$, an integral equation for $a^{-1}\in E$, multiplied by $a^{r-1}$, expresses $a^{-1}$ as an element of $P$. Therefore $P$ must be a field. A polynomial ring in a positive number of variables is not a field, since a variable has no polynomial inverse. Hence $d=0$ and $E/k$ is a finite field extension. This proves the Zariski lemma.

Now let $k$ be an algebraically closed field. The Weak Hilbert Nullstellensatz says that every maximal ideal of $k[X_1,\dots ,X_n]$ is uniquely of the formTo prove it, the residue field $k[X]/\mathfrak{m}$ is a field generated as a $k$-algebra by the images of the variables. The Zariski lemma makes it finite algebraic over $k$, and algebraic closedness makes it $k$. Thus each $X_i$ has an image $a_i\in k$. The evaluation map has kernel the displayed ideal: subtracting the constant value of a polynomial expresses its difference as a combination of $X_i-a_i$. That kernel is maximal and contained in $\mathfrak{m}$, so equality holds. Conversely every evaluation kernel is maximal because its quotient is $k$. Uniqueness follows from the variable images. Every proper ideal is contained in a maximal ideal, so it has a common zero; equivalently, an ideal with no common zero is the whole ring.

For an ideal $I\subseteq k[X_1,\dots ,X_n]$, let $V(I)$ be its common-zero set and let $I(V(I))$ be all polynomials vanishing on that set. The Strong Hilbert Nullstellensatz statesThe inclusion $\sqrt{I}\subseteq I(V(I))$ follows because a field has no nonzero nilpotent elements. For the other inclusion, take $f$ vanishing on $V(I)$, with $f\ne 0$, and form the Rabinowitsch trick idealIt has no common zero: at a zero of $I$ the second generator has value one. The Weak Hilbert Nullstellensatz in $n+1$ variables gives $J=(1)$. Hence a finite identity has the form $1={\sum }_j{h}_j(X,T)g_j(X)+h(X,T)(1-Tf)$, with $g_j\in I$. Substitute $T=f^{-1}$ in the localization of a ring $k[X{]}_f$. Clearing the finitely many powers of $f$ occurring in denominators gives $f^r\in I$ for some $r$, so $f\in \sqrt{I}$. The case $f=0$ is immediate. This completes all three proofs. The algebraically closed hypothesis belongs to the two forms of the Hilbert Nullstellensatz; it was not needed for the Noether normalization lemma or the Zariski lemma.

Let $k\to A$ be a map of commutative rings. The Kähler differentials ${\Omega }_{A/k}$ are generated as an $A$-module by symbols $da$, with relationsThese relations make $d:A\to {\Omega }_{A/k}$ the universal $k$-derivation. Every $k$-derivation $D:A\to M$ into an $A$-module factors uniquely through the map $da\mapsto D(a)$. Thus the universal property of Kähler differentials isThis constructs the module and proves its universal property, rather than merely listing a formal derivative rule.

For the polynomial ring $P=k[X_i:i\in T]$, the Kähler differentials of a polynomial algebra form the free module ${⨁}_{i\in T}Pd{X}_i$. The usual formal partial derivatives prove that assigning arbitrary images to $d{X}_i$ defines a derivation, and every polynomial involves only finitely many variables. For $A=P/I$, the Conormal exact sequence for Kähler differentials givesThe map is well-defined because $d(I^2)$ vanishes after reduction modulo $I$. Its cokernel has exactly the universal property of derivations on $P$ that kill $I$, so is ${\Omega }_{A/k}$. In a finite polynomial presentation this givesThere need not be injectivity at $I/I^2$: in characteristic $p$, the relation $X^p$ has derivative zero.

Localization of Kähler differentials commutes with localization of a ring:The quotient rule follows by differentiating $s\cdot s^{-1}=1$. It extends every derivation uniquely and proves the isomorphism by the universal property. Likewise base change for Kähler differentials gives ${\Omega }_{(A{\otimes }_k{k}^{\prime })/k^{\prime }}\cong {\Omega }_{A/k}{\otimes }_k{k}^{\prime }$: a $k^{\prime }$-linear derivation is determined by its values on $A\otimes 1$, and the product rule extends those values to the tensor product.

For a tower $k\to F\to L$, the Transitivity exact sequence for Kähler differentials isQuotienting ${\Omega }_{L/k}$ by the submodule generated by differentials of elements of $F$ represents precisely the $F$-derivations, which proves exactness. The first map need not be injective in general. To relate this to a transcendence basis, we need the stronger property supplied by a separable field extension.

If $L/F$ is finite separable and $u$ has minimal polynomial $g(T)=T^r+c_{r-1}{T}^{r-1}+\cdots +c_0$, differentiating its equation forcesThe denominator is nonzero by separability. Conversely this formula extends an arbitrary $k$-derivation $F\to M$, for an $L$-module $M$, to $F[u]$; it kills the relation $g$ and hence descends to $L$. A tower of simple separable extensions proves unique extension for all finite separable $L/F$. Therefore Kähler differentials under a separable field extension satisfyThis proves injectivity in this case, which would not follow from right exactness alone.

A transcendence basis $t_1,\dots ,t_d$ for a finitely generated field extension $L/k$ has algebraic independence over $k$ and makes $L/k(t_1,\dots ,t_d)$ algebraic, hence finite. A separating transcendence basis additionally makes that finite extension separable. Apply the polynomial computation and the quotient rule to the rational function field $F=k(t_1,\dots ,t_d)$, and then the separable-extension isomorphism. We obtain the central link:when the basis is separating. The dual statement says that any prescribed values of the $D(t_i)$ extend uniquely to a $k$-derivation of $L$ with values in $L$. In characteristic zero every transcendence basis of a finitely generated field extension is separating, so Kähler differentials measure transcendence degree exactly.

There is also a characteristic-zero test for algebraic independence. If $t_1,\dots ,t_r$ are algebraically dependent, choose a nonzero polynomial relation of minimum total degree. Some formal partial derivative is nonzero in characteristic zero, and it cannot also vanish on the tuple, since it has smaller degree. Differentiation therefore gives a nontrivial linear relation among the $d{t}_i$. Conversely, an algebraically independent tuple extends to a transcendence basis, whose differentials form a basis as just proved. Thus differentials detect algebraic independence in characteristic zero: a finite tuple is algebraically independent exactly when its differentials are linearly independent. Its differentials form a basis of ${\Omega }_{L/k}$ exactly when the tuple is a transcendence basis.

The separability qualification is essential in positive characteristic. For $k={\mathbb{F}}_p(s)$ and $L=k(u)$ with $u^p=s$, the extension is purely inseparable of degree $p$, with transcendence degree zero. The polynomial presentation yields ${\Omega }_{L/k}=Ldu\ne 0$, because the defining relation has derivative zero. Also, in ${\mathbb{F}}_p(t)/{\mathbb{F}}_p$, the tuple $\{t^p\}$ is a transcendence basis but $d(t^p)=0$, while $dt$ is a basis of the one-dimensional differential module. A chosen arbitrary transcendence basis therefore need not give differential coordinates in characteristic $p$.

Finally, the presentation makes the relation useful geometrically. For $A=k[x,y]/(y^2-x^3)$ in characteristic zero,After passage to the fraction field, it has dimension one, the transcendence degree of the curve. At the origin, tensoring with its residue field leaves both $dx,dy$ independent, so the differential fibre has dimension two. For a $k$-rational point with maximal ideal $\mathfrak{m}$, that fibre is the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$: write elements as their constant value plus an element of $\mathfrak{m}$, and note that derivations into $k$ kill ${\mathfrak{m}}^2$. This explains how Kähler differentials record both generic transcendental parameters and the extra tangent direction at a singular point.

In characteristic $p$, let $K={\mathbb{F}}_p(s,t)$ and $L={\mathbb{F}}_p(s^{1/p},t^{1/p})$, with $s,t$ having algebraic independence. The monomials $s^{i/p}{t}^{j/p}$, $0\le i,j<p$, form a $K$-basis, so $[L:K]=p^2$. Their independence follows by grouping polynomial monomials according to their exponents modulo $p$, after clearing rational-function denominators.

Every $\alpha \in L$ has ${\alpha }^p\in K$, so its minimal polynomial divides $T^p-{\alpha }^p$ and $[K(\alpha ):K]\le p$. Thus this finite purely inseparable extension has no primitive element. It demonstrates why the separable field extension hypothesis cannot simply be removed.

Choose a very ample divisor $A$. By the stronger form of Kodaira's lemma, $jD\sim A+E$ for some effective divisor $E$. The subsystem $s_E{H}^0(X,A)\subseteq H^0(X,jD)$ defines the embedding given by $A$ on $X\setminus \mathrm{Supp}E$. Ratios of its sections generate the function field $k(X)$. The map from the complete complete linear system of a divisor contains these ratios, so it induces the same function field and is birational onto its image.

The converse is true; smoothness is unnecessary. If $|jD|$ gives a birational map, choose $n$ algebraically independent ratios $s_1/s_0,\dots ,s_n/s_0$ among a generating set of its section ratios. For every $q$, the sectionsare linearly independent by algebraic independence. Therefore $h^0(X,qjD)\ge \left(\genfrac{}{}{0ex}{}{q+n}{n}\right)$, giving condition (1) of part (iii). This is the birational linear system criterion for bigness.